TU Wien:Analysis UE (diverse)/Übungen SS19/Beispiel 184

Man berechne die Grenzwerte

WS19: (ohne Verwendung der Regel von de l'Hospital):

1. (a) $lim_{x\to 1}\left({\frac {2}{1-x^{2}}}-{\frac {3}{1-x^{3}}}\right)$
2. (b) $lim_{x\to \infty }{\frac {17x^{2}+4x-1}{x^{3}-12x^{2}+1}}$
3. (c) $lim_{x\to 0}\left({\frac {1-cos(x)}{x}}\right)$

Lösungsvorschlag

Beispiel (a)

${\begin{aligned}lim_{x\to 1}\left({\frac {2}{1-x^{2}}}-{\frac {3}{1-x^{3}}}\right)&=lim_{x\to 1}\left({\frac {2}{(1-x)(1+x)}}-{\frac {3}{(1-x)(1+x+x^{2})}}\right)\\&=lim_{x\to 1}\left({\frac {2(1+x+x^{2})-3(1+x)}{(1-x)(1+x)(1+x+x^{2})}}\right)\\&=lim_{x\to 1}\left({\frac {2x^{2}-x-1}{(1-x)(1+x)(1+x+x^{2})}}\right)\\&=lim_{x\to 1}\left({\frac {(2x+1)(x-1)}{(1-x)(1+x)(1+x+x^{2})}}\right)\\&=lim_{x\to 1}\left({\frac {(2x+1)(1-x)(-1)}{(1-x)(1+x)(1+x+x^{2})}}\right)\\&=lim_{x\to 1}\left({\frac {-2x-1}{(1+x)(1+x+x^{2})}}\right)\\&=-{\frac {3}{6}}=-{\frac {1}{2}}\end{aligned}}$

(Das ging auch viel viel einfacher. Habs aber nicht abgeschrieben.)

Beispiel (b)

$lim_{x\to \infty }{\frac {17x^{2}+4x-1}{x^{3}-12x^{2}+1}}=lim_{x\to \infty }{\frac {{\frac {17}{x}}+{\frac {4}{x^{2}}}-{\frac {1}{x^{3}}}}{1-{\frac {12}{x}}+{\frac {1}{x^{3}}}}}=0$

Lösungsvorschlag von Paul&Fabi

Beispiel (c)

${\begin{aligned}lim_{x\to 0}\left({\frac {1-cos(x)}{x}}\right)&=lim_{x\to 0}{\frac {1}{x}}*\left({1-cos(x)}\right)\\&=lim_{x\to 0}{\frac {1}{x}}*\left(1-\left(1-{\frac {x^{2}}{2!}}+{\frac {x^{4}}{4!}}-{\frac {x^{6}}{6!}}+...\right)\right)\\&=lim_{x\to 0}{\frac {1}{x}}*\left(1-1+{\frac {x^{2}}{2!}}-{\frac {x^{4}}{4!}}+{\frac {x^{6}}{6!}}-...\right)\\&=lim_{x\to 0}\left({\frac {x}{2!}}-{\frac {x^{3}}{4!}}+{\frac {x^{5}}{6!}}-...\right)\\&=0\end{aligned}}$